证明0.0999＜1/10"+1/11"+1/12"+.....+1/1000"＜0.111

[1/(10x10)+1/(11x11)+1/(12x12)+.....+1(1000x1000)]>[1/(10x11)+1/(11x12)+1/(12x13)+.....+1(1000x1001)]=(1/10)-(1/11)+(1/11)-(1/12)+.......+(1/1000)-(1/1001)=
(1/10)-(1/1001)=0.099

[1/(10x10)+1/(11x11)+1/(12x12)+.....+1(1000x1000)]<[1/(9x10)+1/(10x11)+1/(11x12)+.....+1(999x1000)]=(1/9)-(1/10)+(1/10)-(1/11)+.......+(1/999)-(1/1000)=
(1/9)-(1/1000)=991/9000=0.110

0.099＜[1/(10x10)+1/(11x11)+1/(12x12)+.....+1(1000x1000)]＜0.110

应该利用以下两式:
1/[(n(n+1)] < 1/n^2 < 1/[(n-1)n]

1/[n(n+1)] = 1/n - 1/(n+1)

则原式 > 1/(10x11)+1/(11x12)+1/(12x13)+.....+1/(1000x1001)

= 1/10 - 1/1001 = 0.099000999000999...

原式 < 1/(9x10)+1/(10x11)+1/(11x12)+.....+1/(999x1000)
= 1/9 - 1/1000 = 0.110111111111..< 0.111

以上可以证明原式< 0.111,但还不能证明原式 > 0.0999.所以要考虑更精确的.先把头尾两项去掉,求出剩余的取值范围,再把头尾两项加回.

1/(11x11)+1/(12x12)+....+1/(999x999) > 1/(11x12)+1/(12x13)+....+1/(999x1000) = 1/11 - 1/